Newton’s Second Law corrected the old belief that force is needed to maintain motion. This idea came from the philosopher ______.
Advertisements
Advertisements
प्रश्न
The monkey B, shown in the following figure, is holding on to the tail of monkey A that is climbing up a rope. The masses of monkeys A and B are 5 kg and 2 kg, respectively. If A can tolerate a tension of 30 N in its tail, what force should it apply on the rope in order to carry monkey B with it? Take g = 10 m/s2.

Advertisements
उत्तर
Let the acceleration of monkey A upwards be a, so that a maximum tension of 30 N is produced in its tail.
T − 5g − 30 − 5a = 0 ...(i)
30 − 2g − 2a = 0 ...(ii)
From equations (i) and (ii), we have:
T = 105 N (max.)
and a = 5 m/s2
So, A can apply a maximum force of 105 N on the rope to carry monkey B with it.
For minimum force, there is no acceleration of A and B.
T1 = weight of monkey B
⇒ T1 = 20 N
Rewriting equation (i) for monkey A, we get:
T − 5g − 20 = 0
⇒ T = 70 N
∴ To carry monkey B with it, monkey A should apply a force of magnitude between 70 N and 105 N.
APPEARS IN
संबंधित प्रश्न
An aircraft executes a horizontal loop at a speed of 720 km/h with its wings banked at 15°. What is the radius of the loop?
A free 238U nucleus kept in a train emits an alpha particle. When the train is stationary, a nucleus decays and a passenger measures that the separation between the alpha particle and the recoiling nucleus becomes x at time t after the decay. If the decay takes place while the train is moving at a uniform velocity v, the distance between the alpha particle and the recoiling nucleus at a time t after the decay, as measured by the passenger, is
In a TV picture tube, electrons are ejected from the cathode with negligible speed and they attain a velocity of 5 × 106 m/s in travelling one centimetre. Assuming straight-line motion, find the constant force exerted on the electrons. The mass of an electron is 9.1 × 10−31 kg.
A man has fallen into a ditch of width d and two of his friends are slowly pulling him out using a light rope and two fixed pulleys as shown in the following figure. Show that the force (assumed equal for both the friends) exerted by each friend on the road increases as the man moves up. Find the force when the man is at a depth h.
Find the mass M of the hanging block in the following figure that will prevent the smaller block from slipping over the triangular block. All the surfaces are frictionless and the strings and the pulleys are light.

Show that the rate of change of momentum = mass × acceleration. Under what condition does this relation hold?
Two bodies A and B of same mass are moving with velocities v and 2v, respectively. Compare their (i) inertia and (ii) momentum.
Two balls A and B of masses m and 2 m are in motion with velocities 2v and v, respectively. Compare:
(i) Their inertia.
(ii) Their momentum.
(iii) The force needed to stop them in the same time.
State Newton's second law of motion. Under what condition does it take the form F = ma?
The unit of linear momentum is :
The linear momentum of a body of mass m moving with velocity v is :
A stone is dropped freely from the top of a tower and it reaches the ground in 4 s. Taking g = 10m s-2, calculate the height of the tower.
A pebble is dropped freely in a well from its top. It takes 20 s for the pebble to reach the water surface in the well. Taking g = 10 m s-2 and speed of sound = 330 m s-1. Find : The time when echo is heard after the pebble is dropped.
A body of mass 200 g is moving with a velocity of 5 ms−1. If the velocity of the body changes to 17 ms−1, calculate the change in linear momentum of the body.
Use Newton's second law to explain the following:
We always prefer to land on sand instead of hard floor while taking a high jump.
A stone is thrown vertically upward with a velocity of 9.8 m/s. When will it reach the ground?
A ball is thrown upward and reaches a maximum height of 19.6 m. Find its initial speed?
The position time graph of a body of mass 2 kg is as given in figure. What is the impulse on the body at t = 0 s and t = 4 s.

As shown in figure, a 70 kg garden roller is pushed with a force of F = 200 N at an angle of 30° with horizontal. The normal reaction on the roller is: (Given: g = 10 ms−2)

